[gpaw-users] Clarification regarding calc.get_xc_difference()

Ask Hjorth Larsen asklarsen at gmail.com
Tue Jan 12 00:42:10 CET 2016


(Please reply to list; re-added)

2016-01-11 15:17 GMT+01:00 Zeeshan Ahmad <2008.zeeshan at gmail.com>:
> Ask,
>
> I think you have misunderstood my question. It was clear that the earlier calculated XC energy was subtracted, the question was regarding what was it subtracted from, if the xc parameter passed was purely correlation functional - like ‘LDA_C_PW’ or ‘GGA_C_PBE’. Will it assume the exchange energy for this purely correlation functional to be zero or take some default xc functional like LDA_X?

It does what you tell it to do, and does not take any strange defaults.

Best regards
Ask

>
> Thanks,
> Zeeshan
>
>> On Jan 11, 2016, at 8:00 AM, Ask Hjorth Larsen <asklarsen at gmail.com> wrote:
>>
>> Dear Zeeshan Ahmad
>>
>> The function always subtracts the XC energy which was already
>> calculated, no matter what xc parameter you pass.
>>
>> If you want something different, I guess you can calculate a
>> difference of two XC differences so the selfconsistently determined XC
>> energy cancels out.
>>
>> (You should be able to verify all this by testing with a little script.)
>>
>> Best regards
>> Ask
>>
>> 2016-01-11 7:41 GMT+01:00 Zeeshan Ahmad <2008.zeeshan at gmail.com>:
>>> Hello GPAW team,
>>>
>>> The gpaw/ase interface has a function called get_xc_difference(xc) which can be used to calculate the difference in the xc energy between the xc functional supplied in the argument of this function and the xc functional which was set for the calculator earlier.
>>>
>>> The question is: if we provide an xc functional (as argument) which handles only correlation part like ‘LDA_C_PW’ or ‘GGA_C_PBE’, then what will get_xc_difference(xc) give?
>>>
>>> Does it calculate the exchange (x) energy using a default functional or assumes the exchange energy = 0. In other words, if the calculator’s functional was set as ‘vdW-DF’ and we do:
>>>
>>> calc.get_xc_difference(‘LDA_C_PW’)
>>>
>>> then, will the result be:
>>>
>>> Difference = E_LDA_correlation - E_exchange_vdW-DF - E_correlation_vdW-DF ?
>>>
>>> (Here the exchange energy of supplied functional was assumed to be zero)
>>>
>>>
>>> Thanks,
>>> Zeeshan Ahmad
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>>> gpaw-users at listserv.fysik.dtu.dk
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>>
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